Inequalities problem

**cakeboby** · Aug 17th 2010, 04:55 PM

Question: Let a and b be the roots of the quadratic equation $x^2- (m +\frac 3 m )x +2 = 0$ where m is a non-zero constant.
Using the fact that $(a-b)^2 \geq 0$ for all real numbers a and b, show that $m^2 +\frac 9 {m}^2 \geq 6$ .
Hence show tha $t (m + \frac 3 m }^2 \geq 12$

MY WORK:
$(a-b)^2 = (a+b)^2 - 4ab = (a+b)^2 - 8 \geq 0$
$(m+ \frac 3 m ) -8 \geq 0$
$m^2+6m+\frac 9 {m}^2 \geq 8$

Thanks in advance.

**Gusbob** · Aug 17th 2010, 07:07 PM

Originally Posted by cakeboby

Question: Let a and b be the roots of the quadratic equation $x^2- (m +\frac 3 m )x +2 = 0$ where m is a non-zero constant.
Using the fact that $(a-b)^2 \geq 0$ for all real numbers a and b, show that $m^2 +\frac 9 {m}^2 \geq 6$ .
Hence show tha $t (m + \frac 3 m }^2 \geq 12$

MY WORK:
$(a-b)^2 = (a+b)^2 - 4ab = (a+b)^2 - 8 \geq 0$
$(m+ \frac 3 m ) -8 \geq 0$
$m^2+6m+\frac 9 {m}^2 \geq 8$

Thanks in advance.

From your first line of working onwards, you'll actually want to use

$(m + \frac{9}{m})^2 - 4(m)(\frac{3}{m}) \geq 0$

$m^2 + 6 + (\frac{9}{m})^2 - 12 \geq 0$

$m^2 + (\frac{9}{m})^2 \geq 6$

Check the binomial expansion. You did this incorrectly in your working.

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